Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{z - 7}{z^2 + z - 56} \div \dfrac{z^2 + z}{z^3 + 18z^2 + 80z} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{z - 7}{z^2 + z - 56} \times \dfrac{z^3 + 18z^2 + 80z}{z^2 + z} $ First factor out any common factors. $q = \dfrac{z - 7}{z^2 + z - 56} \times \dfrac{z(z^2 + 18z + 80)}{z(z + 1)} $ Then factor the quadratic expressions. $q = \dfrac {z - 7} {(z + 8)(z - 7)} \times \dfrac {z(z + 8)(z + 10)} {z(z + 1)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {(z - 7) \times z(z + 8)(z + 10) } { (z + 8)(z - 7) \times z(z + 1)} $ $q = \dfrac {z(z + 8)(z + 10)(z - 7)} {z(z + 8)(z - 7)(z + 1)} $ Notice that $(z + 8)$ and $(z - 7)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {z\cancel{(z + 8)}(z + 10)(z - 7)} {z\cancel{(z + 8)}(z - 7)(z + 1)} $ We are dividing by $z + 8$ , so $z + 8 \neq 0$ Therefore, $z \neq -8$ $q = \dfrac {z\cancel{(z + 8)}(z + 10)\cancel{(z - 7)}} {z\cancel{(z + 8)}\cancel{(z - 7)}(z + 1)} $ We are dividing by $z - 7$ , so $z - 7 \neq 0$ Therefore, $z \neq 7$ $q = \dfrac {z(z + 10)} {z(z + 1)} $ $ q = \dfrac{z + 10}{z + 1}; z \neq -8; z \neq 7 $